Ultraviolet A 716-Commedia dell & amp; #39; arte (3D N digital problem)

Ultraviolet A 716-Commedia dell & #39; arte (3D N digital problem)
Ultraviolet A 716-Commedia dell 'arte

Question Link

Question: Given A Three-Dimensional n digital game, it is required to change to order, and the last position is space. Can it be changed successfully?

Idea: The same as the two-dimensional determination method, because the Z axis moves, it is equal to switching N ^ 2-1 times, the Y axis moves is equal to switching N-1 times, and the X axis moves unchanged, the parity of Reverse Order pairs remains unchanged.

If n is an even number, the reverse order can be an even number, but not an odd number.
When n is an odd number, the Y axis and Z axis of the space position are different in steps and in reverse order. They are the same and can walk differently.

Code:

#include 
 
  #include 
  
   const int N = 1000005;typedef long long ll;int t, n, a[N], save[N], v;ll cal(int *a, int l, int r) {    if (l >= r) return 0;    int mid = (l + r) / 2;    int sl = l, sr = r;    ll ans = cal(a, l, mid) + cal(a, mid + 1, r);    int tmp = mid; mid++;    for (int i = l; i <= r; i++) {if (l <= tmp && mid <= r) {    if (a[l] <= a[mid]) save[i] = a[l++];    else {ans += mid - i;save[i] = a[mid++];    }}else if (l <= tmp) save[i] = a[l++];else if (mid <= r) save[i] = a[mid++];    }    for (int i = sl; i <= sr; i++)a[i] = save[i];    return ans;}bool judge() {    ll nx = cal(a, 0, n * n * n - 2);    if (n&1) {if (nx&1) return false;    }    else {int z[3];for (int i = 0; i < 3; i++) {    z[i] = v % n;    v /= n;}ll bu = 2 * n - 2 - z[2] - z[1];if ((bu^nx)&1) return false;    }    return true;}int main() {    scanf("%d", &t);    while (t--) {scanf("%d", &n);int flag = 0;for (int k = 0; k < n; k++) {    for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {    int now = n * n * k + n * i + j - flag;    scanf("%d", &a[now]);    if (a[now] == 0) {v = now;flag = 1;    }}    }}if (judge()) printf("Puzzle can be solved.\n");else printf("Puzzle is unsolvable.\n");    }    return 0;}